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nydimaria [60]
3 years ago
9

During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 0.88

8 km/s at an initial inclination of 46.1° to the horizontal. The acceleration of gravity is 9.8 m/s 2. How far away did the shell hit?
Physics
1 answer:
Brut [27]3 years ago
4 0

369.38 km

speed,x = 2.07cos(28.8°) = 1.81 km/s

speed,y = 2.07sin(28.8°) = 1.00 km/s

v_{f}=v_{0} +a * t

0 = 1000 - 9.8 * t

9.8 * t = 1000

t = 102.04 s

After reaching the highest point the shell takes the same time to reach the ground where it was fired, so the total time it flies is 102.04* 2 = 204.08 s

d = 1.81 km/s * 204.08 s (I used the speed in km/s because the answer needs to be in km)

d = 369.38 km

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You have about 10 quarts of blood in your body. At REST your heart pumps about 5 quarts each minutes. That is half of your blood
Zina [86]

Answer:

8 times

Explanation:

Given that You have about 10 quarts of blood in your body. At REST your heart pumps about 5 quarts each minutes.

That means the heart will pump 10 quarts in 2 minutes.

That is half of your blood volume per minute.

If during exercise it can pump 40 quarts per minute, that is, 80 quarts in 2 minutes.

To know how many times does all of your blood complete the cycle around your body during exercise, you must divide 80 quarts by 10 quarts. That is,

80 / 10 = 8

Therefore, your blood complete the cycle around your body 8 times during the exercise.

3 0
3 years ago
A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and
iogann1982 [59]

Answer:

p_2 = 1.76 atm

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}

Putting the values

\frac{1.15*2.37}{280}  =\frac{p_2 *1.68}{304}

9.733*10^{-3} = \frac{p_2 *1.68}{304}

9.733*10^{-3}*304 = p_2*1.68

\frac{9.733*10^{-3}*304}{1.68} =p_2

p_2= 1.76 atm

7 0
3 years ago
A small mailbag is released from a helicopter that is descending steadily at 3 m/s.
mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

       Speed of mailbag after 3 seconds = 26.4 m/s

       Package is 44.1 meter below helicopter

<u>Explanation:</u>

a)  We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

   Initial velocity = 3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

   v = 3+9.8*3 = 32.4 m/s

  Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m

  Distance traveled by package  = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

  v = -3+9.8*3 = 26.4 m/s

  Speed of mailbag after 3 seconds = 26.4 m/s

 Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m

  Distance traveled by package  = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0
3 years ago
1. A rock of granite has a mass of 50 kg. if it’s weight in water
igor_vitrenko [27]

Answer:

The first part of the question is asking about BUOYANT FORCE or UPTHRUST.

Upthrust =TRUE WEIGHT-APPARENT WEIGHT

TRUE WEIGHT=mg

TRUE weight=50kg×10m/s²

=500N

upthrust=500N-380N

FB=120N

volume of the rock=mass/density.

since the granite is completely submerged, the volume of the displaced liquid will be equal to the volume of the body.

upthrust=Vdg

120N=V×1000kg/m³×10m/s²

120N=V×10000kg/m²s²

120/10000=V

v=0.012m³

please mark brainliest, hope it helped

6 0
2 years ago
A 48 kg cart is sitting motionless at the top of the hill. At what height, did the cart start at, to reach 88435 J of energy at
svet-max [94.6K]

Answer : The height is 188 meters  

Explanation :   When the cart reached at the end from top of hill then the cart have potential energy .

Given that,

Potential energy = 88435 J

Mass of cart = 48 kg

We know that,

The potential energy is

mgh =88435

h= \dfrac{88435}{48\times9.8}

h = 187.9 = 188\ meters

So, the height of the top is 188 meters.

7 0
3 years ago
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